How do you write the quadratic in vertex form given #y=-4x^2-5x+3#?

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Answer 1 · 2015-05-05T00:28:15

#y= (-4)(x+5/8)^2 + 73/16#

The general vertex form of a quadratic is
#y = m(x-a)^2+b#
where #(a,b)# is the vertex

#y= -4x^2-5x +3#

#y= (-4)(x^2+5/4x) +3 " extract the "m" factor"#

#y=(-4)(x^2+5/4x+(5/8)^2) -(-4)(5/8)^2 + 3#

#y= (-4)(x+5/8)^2 + 73/16#

...and (assuming I haven't made any mistakes) the vertex is at #(-5/8,73/16)=(-5/8,4 9/16)#

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